NCERT Solutions for Class 12 Physics Chapter 4 Moving Charges and Magnetism is given here for the benefit of CBSE class science students. In chapter 4 of class 12 NCERT physics book, students are given knowledge on the concepts of moving charges and magnetism. In this chapter several topics with good marks allotment like magnetic field and direction of a circular coil, magnetic force, etc. are covered. These concepts are not only important for the CBSE board exam but also are crucial for the competitive exams like JEE and NEET.

Moving charges and magnetism NCERT solutions pdf provided here has answers to textbook questions along with class 12 physics chapter 4 important questions from previous year question papers, exercises and worksheets that help you in preparing chapter 4 physics class 12 notes to keep it handy when the exam nears.

## Class 12 Physics NCERT Solutions for Moving Charges and Magnetism

NCERT class 12 physics solutions for chapter 4 moving charges and magnetism given here can help the students to clear all their doubts in an instant. This chapter comprises of diagrams, graphs, illustrations and day to day examples that make your study interesting and helps to remember the concepts for a longer period.

The direct derivations of topics like forces between two current carrying wires, ampere circuital law are frequently asked in exams. Numerical problems are asked from this chapter. The working and figure of cyclotron should be studied intensively.

### Subtopics of NCERT Class 12 Physics Chapter 4 Moving Charges and Magnetism

Section Number |
Topic |

4.1 | Introduction |

4.2 | Magnetic Force |

4.2.1 | Sources And Fields |

4.2.2 | Magnetic Field, Lorentz Force |

4.2.3 | Magnetic Force On A Current-carrying Conductor |

4.3 | Motion In A Magnetic Field |

4.4 | Motion In Combined Electric And Magnetic Fields |

4.4.1 | Velocity Selector |

4.4.2 | Cyclotron |

4.5 | Magnetic Field Due To A Current Element, Biot-savart Law |

4.6 | Magnetic Field On The Axis Of A Circular Current Loop |

4.7 | Ampere’s Circuital Law |

4.8 | The Solenoid And The Toroid |

4.8.1 | The Solenoid |

4.8.2 | The Toroid |

4.9 | Force Between Two Parallel Currents, The Ampere |

4.10 | Torque On Current Loop, Magnetic Dipole |

4.10.1 | Torque On A Rectangular Current Loop In A Uniform Magnetic Field |

4.10.2 | Circular Current Loop As A Magnetic Dipole |

4.10.3 | The Magnetic Dipole Moment Of A Revolving Electron |

4.11 | The Moving Coil Galvanometer |

### Class 12 Physics NCERT Solutions Moving Charges and Magnetism Important Questions

**Q 4.1) Determine the magnitude of the magnetic field B at the centre of the circular coil of wire carrying current of 0.4 A and having 100 turns with 8 cm being the radius of each turn. **

**Answer 4.1:**

Number of turns on the coil (n) = 100

Radius of each turn (r) = 8 cm = 0.08 m

Current flowing in the coil (I) = 0.4 A

The following relation gives the magnitude of the magnetic field at the centre of the coil:

\(|\bar B| = \frac{\mu_{0}\; 2\pi nl}{4\pi r}\)Where,

\(\mu_{0}\) = Permeability of free space= \(4\pi \times 10^{-7}\; T\;m\;A^{-1}\)

So,

\(|\bar B| = \frac{4\pi \times 10^{-7}}{4\pi }\times \frac{2\pi \times 100\times 0.4}{r}\)= \(3.14 \times 10^{-4}\; T\)

Hence, the magnitude of the magnetic field is \(3.14 \times 10^{-4}\; T\).

**Q 4.2) Determine the magnitude of the field B at a point that is 20 cm away from the wire through which 35 A current flows.**

**Answer 4.2:**

Current in the wire (I) = 35 A

Distance of a point from the wire (r) = 20 cm = 0.2 m

At this point the magnitude of the magnetic field is given as:

\(|\bar B| = \frac{\mu_{0}\; 2l}{4\pi\; r}\)Where,

\(\mu_{0}\) = Permeability of free space= \(4\pi \times 10^{-7}\; T\;m\;A^{-1}\)

So,

\(|\bar B| = \frac{4\pi \times 10^{-7}}{4\pi }\times \frac{2\times 35}{0.2}\)= \(3.5 \times 10^{-5}\; T\)

Hence, the magnitude of the magnetic field at a point 20 cm from the wire is \(3.5 \times 10^{-5}\; T\).

**Q 4.3) Determine the direction and magnitude of B at a point that is 2.5 m away in the east direction of the long straight wire that is in a horizontal plane carrying a current of 50 A in North to South direction.**

**Answer 4.3:**

Current in the wire (I) = 50 A

A point is 2.5 m away from the East of the wire.

Therefore, Magnitude of the distance of the point from the wire (r) = 2.5 m

Magnitude of the magnetic field at that point is given by the relation:

\(|\bar B| = \frac{\mu_{0}\; 2l}{4\pi\; r}\)Where,

\(\mu_{0}\) = Permeability of free space= \(4\pi \times 10^{-7}\; T\;m\;A^{-1}\)

\(|\bar B| = \frac{4\pi \times 10^{-7}}{4\pi }\times \frac{2\times 50}{2.5}\)

= \(4 \times 10^{-6}\; T\)

The point is located normal to the wire length at a distance of 2.5 m. The direction of the current in the wire is vertically downward. Hence, according to the Maxwell’s right hand thumb rule, the direction of the magnetic field at the given point is vertically upward.

**Q 4.4) A flat overhead electrical cable carries a current of 90 A in east to west course. What is the direction and magnitude of the magnetic field due to the current 1.5 m below the line?**

**Answer 4.4:**

Current in the power line (I) = 90 A

Point is located below the electrical cable at distance (r) = 1.5 m

Hence, magnetic field at that point is given by the relation,

\(|\bar B| = \frac{\mu_{0}\; 2l}{4\pi\; r}\)Where,

\(\mu_{0}\) = Permeability of free space= \(4\pi \times 10^{-7}\; T\;m\;A^{-1}\)

\(|\bar B| = \frac{4\pi \times 10^{-7}}{4\pi }\times \frac{2\times 90}{1.5}\)

= \(1.2 \times 10^{-5}\; T\)

The current is flowing from East to West. The point is below the electrical cable.

Hence, according to Maxwell’s right hand thumb rule, the direction of the magnetic field is towards the South.

** **

**Q 4.5) A wire carrying a current of 8 A makes an angle of 30 ^{0} with the direction of a uniform magnetic field of 0.15 T. Find the magnitude of magnetic force per unit length on the wire.**

**Answer 4.5:**

Current in the wire (I) = 8 A

Magnitude of the uniform magnetic field (B) = 0.15 T

Angle between the wire and magnetic field, \(\theta = 30°\).

Magnetic force per unit length on the wire is given as: F = \(BI sin\theta\)

= \(0.15\times 8\times 1\times sin 30°\)

= \(0.6\; N\; m^{-1}\)

Hence, the magnetic force per unit length on the wire is \(0.6\; N\; m^{-1}\).

**Q 4.6) Determine the magnetic force on a wire of 3 cm, carrying a current of 10 A and placed inside a solenoid normal to its axis. Inside the solenoid, the magnetic field is given as 0.27 T.**

**Answer 4.6:**

Length of the wire (l) = 3 cm = 0.03 m

Current flowing in the wire (I) = 10 A

Magnetic field (B) = 0.27 T

Angle between the current and magnetic field, \(\theta = 90°\).

Magnetic force exerted on the wire is given as:

\(F = BIl sin \theta\)= \(0.27\times 10\times 0.03\; sin90°\)

= \(8.1\times 10^{-2}\; N\)

Hence, the magnetic force on the wire is \(8.1\times 10^{-2}\; N\). The direction of the force can be obtained from Fleming’s left hand rule.

** **

**Q 4.7) Two long and parallel wires, wire A carrying current 8.0 A and wire B carrying current 5.0 A in the same direction are isolated by a distance of 4 cm. Calculate the force on a 10 cm section of wire A.**

**Answer 4.7:**

Current flowing in wire A (\(I_{A}\)) = 8 A

Current flowing in wire B (\(I_{B}\)) = 5 A

Distance between the two wires (r) = 4 cm = 0.04 m

Length of a section of wire A (L) = 10 cm = 0.1 m

Force exerted on length L due to the magnetic field is given as:

\(F = \frac{\mu_{o}I_{A}I_{B}L}{2\pi r}\)Where,

\(\mu_{0}\) = Permeability of free space = \(4\pi \times 10^{-7}\; T\;m\;A^{-1}\) \(F = \frac{4\pi \times 10^{-7}\times 8\times 5\times 0.1}{2\pi\times 0.04} = 2\times 10^{-5} \;N\)The magnitude of force is \(2\times 10^{-5} \;N\). This is an attractive force normal to A towards B because the direction of the currents in the wires is the same.

**Q 4.8) Find the magnitude of magnetic field B inside the solenoid carrying current of 8.0 A near its centre. Given Solenoid diameter as 1.8 cm, length as 80 cm and has 5 layers of windings of 400 turns each.**

**Answer 4.8:**

Solenoid length (l) = 80 cm = 0.8 m

Five layers of windings of 400 turn each on the solenoid.

\(∴\) Total number of turns on the solenoid, N = 5 x 400 = 2000Solenoid Diameter (D) = 1.8 cm = 0.018 m

Current carried by the solenoid (I) = 8.0 A

The relation that gives the magnitude of magnetic field inside the solenoid near its centre is given below:

\(B = \frac{\mu_{0}NI}{l}\)Where,

\(\mu_{0}\) = Permeability of free space = \(4\pi \times 10^{-7}\; T\;m\;A^{-1}\) \(B = \frac{4\pi \times 10^{-7}\times 2000\times 8}{0.8}\)= \(2.5\times 10^{-2}\; T\)

Hence, The magnitude of B inside the solenoid near its centre is \(2.5\times 10^{-2}\; T\).

**Q 4.9) Determine the magnitude of the torque experienced by the square coil when it is suspended vertically and the normal to the plane makes an angle of 30 ^{0} with the direction of a uniform horizontal magnetic field of magnitude 0.8 T. Taking square coil of side 10 cm consisting of 20 turns and carrying a current of 12 A.**

**Answer 4.9:**

Length of a side of the square coil (l) = 10 cm = 0.1 m

Current flowing in the coil (I) = 12 A

Number of turns on the coil (n) = 20

Angle made by the plane of the coil with B (Magnetic field), \(\theta = 30°\)

Strength of magnetic field, B = 0.8 T

The following relation gives the magnitude of the magnetic torque experienced by the coil in the magnetic field:

\(\tau = n\; BIA\; sin\theta\)Where,

A = Area of the square coil

= 1 x 1 = 0.1 x 0.1 = 0.01 m^{2}

So, \(\tau = 20\times 0.8\times 12\times 0.01\times sin30°\)

= 0.96 N m

Hence, 0.96 N m is the magnitude of the torque experienced by the coil.

**Q 4.10) Find the ratio of **

**(a) Current sensitivity of M _{2 }and M_{1}**

**(b) Voltage sensitivity of M _{2 }and M_{1}**

**The moving coil meters, M _{1 }and M_{2} have the following particulars:**

**\(R_{1} = 10\; \Omega ,\; N_{1} = 30\),**

**\(A_{1} = 3.6\times 10^{-3}\; m^{2},\; B_{1} = 0.25\; T\)**

**\(R_{2} = 14\; \Omega ,\; N_{2} = 42\),**

**\(A_{2} = 1.8\times 10^{-3}\; m^{2},\; B_{2} = 0.5\; T\)**

**(Note: Spring constants are identical for the two meters).**

**Answer 4.10:**

Given data:

Moving coil meter M_{1} Moving coil meter M_{2}

Resistance, \(R_{1} = 10\; \Omega\) Resistance, \(R_{1} = 10\; \Omega\)

Number of turns, \(N_{1} = 30\) Number of turns, \(N_{2} = 42\)

Area, \(A_{1}=3.6\times 10^{-3}\;m^{2}\) Area, \(A_{2}=1.8\times 10^{-3}\;m^{2}\)

Magnetic field strength, B_{1} = 0.25 T Magnetic field strength, B_{2} = 0.5 T

Spring constant K_{1} = K Spring constant K_{2} = K

(a) Current sensitivity of M_{1} is given as:

And, Current sensitivity of M_{2} is given as:

Hence, the ratio of current sensitivity of M_{2} to M_{1} is 1.4.

(b) Voltage sensitivity for M_{2} is given as:

And, voltage sensitivity for M_{1} is given as:

Hence, the ratio of voltage sensitivity of M_{2} to M_{1} is 1.

**Q 4.11) In a chamber, a uniform magnetic field of 6.5 G (1 G = 10 ^{-4} T) is maintained. An electron is shot into the field with a speed of 4.8 x 10^{6} m s^{-1} normal to the field. Explain why the path of the electron is a circle. Determine the radius of the circular orbit. (\(e=1.6\times 10^{-19} \;C,\; m_{e}=9.1\times 10^{-31} \;kg\))**

**Answer 4.11:**

Magnetic field strength (B) = 6.5 G = \(6.5\times 10^{-4}\; T\)

Speed of the electron (v) = \(4.8\times 10^{6}\; m/s\)

Charge on the electron (e) = \(1.6\times 10^{-19}\; C\)

Mass of the electron (m_{e}) = \(9.1\times 10^{-31}\; kg\)

Angle between the shot electron and magnetic field, \(\theta = 90°\)

The relation for Magnetic force exerted on the electron in the magnetic field is given as:

\(F = evB\; sin\theta\)This force provides centripetal force to the moving electron. Hence, the electron starts moving in a circular path of radius r.

Hence, centripetal force exerted on the electron,

\(F_{e} = \frac{mv^{2}}{r}\)In equilibrium, the centripetal force exerted on the electron is equal to the magnetic force i.e.,

F_{e} = F

=> \(\frac{mv^{2}}{r} = evB\; sin\theta\)

=> \(r = \frac{mv}{eB\;sin\theta}\)

So,

\(r = \frac{9.1\times 10^{-31}\times 4.8\times 10^{6}}{6.5\times 10^{-4}\times 1.6\times 10^{-19}\times sin90°} = 4.2\times 10^{-2}\; m = 4.2 cm\)Hence, 4.2 cm is the radius of the circular orbit of the electron.

**Q 4.12) In Exercise 4.11 find the frequency of revolution of the electron in its circular orbit. Does the answer depend on the speed of the electron? Explain.**

**Answer 4.12:**

Magnetic field strength (B) = \(6.5\times 10^{-4}\; T\)

Charge on the electron (e) = \(1.6\times 10^{-19}\; C\)

Mass of the electron (m_{e}) = \(9.1\times 10^{-31}\; kg\)

Speed of the electron (v) = \(4.8\times 10^{6}\; m/s\)

Radius of the orbit, r = 4.2 cm = 0.042 m

Frequency of revolution of the electron = v

Angular frequency of the electron = \(\omega = 2\pi v\)

Velocity of the electron is related to the angular frequency as: \(v = r\omega\)

In the circular orbit, the magnetic force on the electron is balanced by the centripetal force.

Hence, we can write:

\(\frac{mv^{2}}{r} = evB\)=> \(eB = \frac{mv}{r} = \frac{m(r\omega)}{r} = \frac{m(r.2\pi v)}{r}\)

=> \(v = \frac{Be}{2\pi m}\)

This expression for frequency is independent of the speed of the electron. On substituting the known values in this expression, we get the frequency as:

\(v = \frac{6.5\times 10^{-4}\times 1.6\times 10^{-19}}{2\times 3.14\times 9.1\times 10^{-31}} = 1.82\times 10^{6}\; Hz \approx 18\;MHz\)Hence, the frequency of the electron is around 18 MHz and is independent of the speed of the electron.

**Q 4.13) (a) A circular coil having radius as 8.0 cm, number of turn as 30 and carrying a current of 6.0 A is suspended vertically in a uniform horizontal magnetic field of magnitude 1.0 T. The field lines make an angle of 60 ^{0} with the normal of the coil. To prevent the coil from turning, determine the magnitude of the counter torque that must be applied.**

**(b) Would your answer change, if the circular coil in (a) were replaced by a planar coil of some irregular shape that encloses the same area? (All other particulars are also unaltered.)**

**Answer 4.13:**

(a) Number of turns on the circular coil (n) = 30

Radius of the coil (r) = 8.0 cm = 0.08 m

Area of the coil = \(\pi r^{2} = \pi (0.08)^{2} = 0.0201 \;m^{2}\)

Current flowing in the coil (I) = 6.0 A

Magnetic field strength, B = 1 T

Angle between the field lines and normal with the coil surface, \(\theta = 60°\)

The coil experiences a torque in the magnetic field. Hence, it turns. The counter torque applied to prevent the coil from turning is given by the relation,

\(\tau = nIBA\; sin\theta\)= \(30\times 6\times 1\times 0.0201\times sin60°\)

= 3.133 N m

(b) It can be inferred from the relation \(\tau = nIBA\; sin\theta\) that the magnitude of the applied torque is not dependent on the shape of the coil. It depends on the area of the coil. Hence, the answer would not change if the circular coil in the above case is replaced by a planar coil of some irregular shape that encloses the same area.

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